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Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.

Squaring an odd-length permutation cycle retains its length, whereas an even-length cycle splits into two cycles of half its length. Thus if any even cycle is present, the number of cycles strictly increases, so the cycle type changes. Since conjugate permutations in a symmetric group have the same cycle type, a permutation is conjugate to its square precisely when all its cycle lengths are odd, equivalently when its order of a group element is odd.

A permutation cycle $(a_1{a}_2\dots a_k)$ sends $a_i$ to $a_{i+1}$, sends $a_k$ to $a_1$, and fixes every other element. Its order is $k$.

Conjugation preserves order of a group element, because $(gx{g}^{-1}{)}^k=g{x}^k{g}^{-1}$, and preserves the sign of a permutation, because sign is a group homomorphism into $\{1,-1\}$. These invariants do not determine cycle type. In $S_6$, $(123)$ and $(123)(456)$ both have order three and positive sign, but have three and zero fixed points respectively. Conjugation relabels cycles, so these permutations are not conjugate.

A linear operator $T:V\to V$ on a vector space over a field $k$ makes $V$ a module over the polynomial ring $k[t]$: let $t$ act as $T$, let constants act as the original scalars, and define $p(t)\cdot v=p(T)v$. The identity $(pq)(T)=p(T)q(T)$ for polynomials $p,q$ verifies the multiplication axiom. The submodules are precisely the invariant subspaces of $T$. This construction is described in Keith Conrad's Introductory Notes on Modules, section 6.

For a computed example, take $k=\mathbb{Q}$, $V=k^2$, andThen $T{e}_2=e_1$, $T{e}_1=0$, and $T^2=0$. ConsequentlyEvery $(a,b)$ equals $(b+at)\cdot e_2$. The module homomorphism $k[t]\to V$, $p\mapsto p(T)e_2$, is therefore surjective, and its kernel is the ideal $(t^2)$. It induces a module isomorphism $V\cong k[t]/(t^2)$. This is a cyclic module and a torsion module over $k[t]$, since $t^{2}V=0$. Over $k$ it is a free module with basis $(e_1,e_2)$ and has no nonzero torsion elements of a module. Over $\mathbb{Q}$ its nonzero elements also have infinite additive order of a group element. Changing scalars changes the questions one asks about the same addition law.

Let $R$ be a complete discrete valuation ring of residue characteristic of a field $p$, with maximal ideal $\mathfrak{m}$, and let $F$ be a one-dimensional formal group law over $R$. Its multiplication series is $[n{]}_F(T)=nT+O(T^2)$. If $p\nmid n$, the invertible morphism criterion for formal group laws constructs its inverse with coefficients in $R$. Both formal power series converge on $\mathfrak{m}$, since their coefficients are integral and the powers of an element of $\mathfrak{m}$ tend to zero. Their compositions are the identity there. Thus multiplication by $n$ is bijective, including when $p=2$ and $n$ is odd. Every finite-order element of $F(\mathfrak{m})$ consequently has order of a group element equal to a prime power with prime $p$.

For a module $M$ over an integral domain $R$, a torsion element is an $m\in M$ with $rm=0$ for some $0\ne r\in R$. This includes zero. Equivalently, its annihilator ${\mathrm{Ann}}_R(m)=\{r\in R:rm=0\}$ is a nonzero ideal. The torsion elements of a module form the torsion submodule.

For $M=\mathbb{Z}/6\mathbb{Z}$, computed annihilators areFor these three elements, the additive order of a group element is respectively $6$, $3$, and $2$: these are the smallest positive integers killing them. Torsion over $\mathbb{Z}$ agrees exactly with finite additive order of a group element. Over a general integral domain, torsion concerns the chosen scalars instead; the polynomial module from a linear operator gives a torsion example whose nonzero elements have infinite additive order of a group element.

A torsion-free group is a group in which the identity element is the only element of finite order of a group element.

Let $E/{\mathbb{Q}}_p$ have good reduction, and choose an integer $m>1/(p-1)$. Its filtration of elliptic-curve points over a local field has a torsion-free group $E_m$ identified by the formal logarithm with $p^m{\mathbb{Z}}_p$. The quotient group $G=E({\mathbb{Q}}_p)/E_m$ is finite. For $g\in G$ of order of a group element $N$, choose a lift $P$. There is a torsion point of an elliptic curve in this coset exactly when the displayed condition holds. Indeed, it is equivalent to the existence of $Q\in E_m$ with $[N]Q=[N]P$. Then $P-Q$ is killed by $N$; conversely every torsion representative has order $N$, since $E_m$ is torsion-free. The logarithmic division and the formal group exponential give the unique such $Q$. Changing the lift changes the logarithm by an element of $N{p}^m{\mathbb{Z}}_p$, so the criterion is well-defined. This finite-coset test includes residue-characteristic torsion; only the prime-to-p torsion lifts uniquely at good reduction are automatic.

A nonidentity rational point in the formal kernel of a minimal Weierstrass equation at two distinct primes has infinite order of a group element. Indeed, the prime-to-residue-characteristic multiplication on a formal group implies that a finite order in the kernel at $p$ must be a power of $p$. Membership in the kernel at a different prime $q$ also forces a power of $q$. The only common possibility is order one, contradicting nonidentity. For an integral model, negative valuations of the affine coordinates provide a convenient kernel-membership test.

For every positive integer $n$, some imaginary quadratic field has an ideal class of order of a group element greater than $n$. Specifically, take a prime number $p\equiv 1(\mathrm{mod}20)$ with $p>5^n$. In $\mathbb{Z}[\sqrt{-p}]$, the Dedekind factorization theorem gives $(5)=P\overline{P}$, with distinct prime ideals $P=(5,\sqrt{-p}-2)$ and $\overline{P}=(5,\sqrt{-p}+2)$ of ideal norm five. If $P^k=(a+b\sqrt{-p})$ for $1\le k\le n$, the field norm equation $5^k=a^2+p{b}^2$ forces $b=0$. Then $k$ is even and $(a)=(5{)}^{k/2}=P^{k/2}{\overline{P}}^{k/2}$, contradicting unique factorization of ideals in a number field. Thus $[P]$ has order of a group element greater than $n$. The existence of suitable $p$ follows from Dirichlet's theorem on arithmetic progressions.